Question: Evaluate $\dfrac{d}{dx}\sqrt[4]{x^3+4x^2+7}$ at $x=-3$. Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $16$ (Choice C) C $\dfrac1{32}$ (Choice D) D $\dfrac3{32}$
Let's start by finding the expression for $\dfrac{d}{dx}\sqrt[4]{x^3+4x^2+7}$. Then, we can evaluate it at $x=-3$. $\sqrt[4]{x^3+4x^2+7}$ is a radical expression, but its argument isn't simply $x$. Therefore, it defines a composite radical function. In other words, suppose $u(x)=x^3+4x^2+7$, then $\sqrt[4]{x^3+4x^2+7}=\sqrt[4]{u(x)}$. $\dfrac{d}{dx}\sqrt[4]{x^3+4x^2+7}$ can be found using the following identity: $\dfrac{d}{dx}\sqrt[4]{u(x)}=\dfrac{1}{4}[u(x)]^{^{-\scriptsize\dfrac{3}{4}}}u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\sqrt[4]{x^3+4x^2+7} \\\\ &=\dfrac{d}{dx}\sqrt[4]{u(x)}&&\gray{\text{Let }u(x)=x^3+4x^2+7} \\\\ &=\dfrac{1}{4}[u(x)]^{^{-\scriptsize\dfrac{3}{4}}}u'(x) \\\\ &=\dfrac{1}{4}[x^3+4x^2+7]^{^{-\scriptsize\dfrac{3}{4}}}(3x^2+8x)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ Now let's evaluate $\dfrac{d}{dx}\sqrt[4]{x^3+4x^2+7}$ at $x= -3$. $\begin{aligned} &\phantom{=}\dfrac{1}{4}\Bigl(({-3})^3+4( {-3})^2+7\Bigr)^{^{-\scriptsize\dfrac{3}{4}}}\cdot\Bigl(3({-3})^2+8({-3})\Bigr) \\\\ &=\dfrac{1}{4}\cdot 16^{^{-\scriptsize\dfrac{3}{4}}}\cdot(3) \\\\ &=\dfrac{1}{4}\cdot \dfrac18\cdot (3) \\\\ &=\dfrac3{32} \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\sqrt[4]{x^3+4x^2+7}$ at $x=-3$ is $\dfrac3{32}$.